1.建立web工程
2.实现servlet
package lin;
import javax.servlet.*;
import javax.servlet.http.*;
import java.io.*;
public class servlet extends GenericServlet{
public void service(ServletRequest request,ServletResponse response)throws IOException , ServletException{
response.setCharacterEncoding("GBK");
PrintWriter out = response.getWriter();
out.println("<html>");
out.println("<head>");
out.println("<title>Servlet 简单例</title>");
out.println("</head>");
out.println("<body>");
out.println("<center>");
out.println("<h2>第Servlet例</h2>");
out.println("</center>");
out.println("</body>");
out.println("</html>");
out.close();
}
}
3.工程web.xml配置servlet访问url
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.0"
<servlet>
<servlet-name>ServletSample</servlet-name>
<servlet-class>lin.servlet</servlet-class>
</servlet>
<!-- 配置Servlet映射路径 -->
<servlet-mapping>
<servlet-name>ServletSample</servlet-name>
<url-pattern>/servlet</url-pattern>
</servlet-mapping>
</web-app>
4.发布工程tomcat
5.启tomcat
6.locatlhost:8080/工程名/servlet路径
例我给例运行：localhost:8888/SampleServlet/servlet
补充：
1、xm<servlet-name></servlet-name>其间值要能相同
2. MyEclipse 8.5 自配置web.xml文件需注意查看xml文件servlet访问路径
2.实现servlet
package lin;
import javax.servlet.*;
import javax.servlet.http.*;
import java.io.*;
public class servlet extends GenericServlet{
public void service(ServletRequest request,ServletResponse response)throws IOException , ServletException{
response.setCharacterEncoding("GBK");
PrintWriter out = response.getWriter();
out.println("<html>");
out.println("<head>");
out.println("<title>Servlet 简单例</title>");
out.println("</head>");
out.println("<body>");
out.println("<center>");
out.println("<h2>第Servlet例</h2>");
out.println("</center>");
out.println("</body>");
out.println("</html>");
out.close();
}
}
3.工程web.xml配置servlet访问url
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.0"
<servlet>
<servlet-name>ServletSample</servlet-name>
<servlet-class>lin.servlet</servlet-class>
</servlet>
<!-- 配置Servlet映射路径 -->
<servlet-mapping>
<servlet-name>ServletSample</servlet-name>
<url-pattern>/servlet</url-pattern>
</servlet-mapping>
</web-app>
4.发布工程tomcat
5.启tomcat
6.locatlhost:8080/工程名/servlet路径
例我给例运行：localhost:8888/SampleServlet/servlet
补充：
1、xm<servlet-name></servlet-name>其间值要能相同
2. MyEclipse 8.5 自配置web.xml文件需注意查看xml文件servlet访问路径
温馨提示：内容为网友见解，仅供参考
第1个回答 2017-09-20
Web程序要通过tomcat启动的，不用eclispe不行的。
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